Please just show me the partial fraction expression and how to get the expression for h[n]. I don't know how to get the expression of h[n].
The above expression simplifies to
So initialize b = [1,-0.5,-1,0.5] and a = [1,-90.6,0.08,0]
Note for no linear constant term in the denominator we still had to define the coefficients as 0.
These arrays define the coefficients of various powers of z in numerator and denominator respectively, the defination is from the highest to the lowest power of z in the expression.
Now using the matlab function of residue we can decompose the this fraction into partial fractions.
[r,p,k] = residuez(b,a)
The r matrix gives you the numerator and the correspondiing terms in p give you the poles for the corresponding partial fractions. k is just is just a multiplicative constant which might be a polynomial.
Example for reference:
Find the partial fraction expansion of the following ratio of polynomials F(z) using residuez
F=b(z)/a(z)=(−4z+8)/(z^{2}+6z+8).
b = [-4 8]; a = [1 6 8]; [r,p,k] = residue(b,a)
r = 2×1 -12 8
p = 2×1 -4 -2
k = []
This represents the partial fraction expansion
(−4z+8)/(z^{2}+6z+8)=(−12/(z+4))+(8/(z+2.))
Convert the partial fraction expansion back to polynomial coefficients using residue.
Note: This example is taken from the documentation website for matlab I would encourage you to look at more examples there for better understanding.
Similarly iztrans function can be used to get back to the discrete time domain form for the transfer function, you can use the following script to get to the discrete time domain defination or h[n]:
syms z n
F = (z^3-0.5*z^2-z+0.5)/(z^3-90.6*z^2+0.08*z)
E = iztrans(F)
doublen = double(n) %this converts the symbolic variable n to double i.e. now you can use doublen plot the response from n= 0 to 10 using doublen.
Please just show me the partial fraction expression and how to get the expression for h[n]. I don't know how to get...
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